The Stacks project

Proof. If $I \subset R$ is an ideal, then the short exact sequence $0 \to I \to R \to R/I \to 0$ gives an exact sequence

by Algebra, Lemma 10.71.7 and the fact that $\mathop{\mathrm{Ext}}\nolimits ^1_ R(R, J) = 0$ as $R$ is projective (Algebra, Lemma 10.77.2). Thus (2) and (3) are equivalent. In this proof we will show that (1) $\Leftrightarrow $ (3) which is known as Baer's criterion.

Assume (1). Given a module map $I \to J$ as in (3) we find the extension $R \to J$ because the map $\mathop{\mathrm{Hom}}\nolimits _ R(R, J) \to \mathop{\mathrm{Hom}}\nolimits _ R(I, J)$ is surjective by definition.

Assume (3). Let $M \subset N$ be an inclusion of $R$-modules. Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi $ extends to $N$ which finishes the proof of the lemma. Consider the set of homomorphisms $\varphi ' : M' \to J$ with $M \subset M' \subset N$ and $\varphi '|_ M = \varphi $. Define $(M', \varphi ') \geq (M'', \varphi '')$ if and only if $M' \supset M''$ and $\varphi '|_{M''} = \varphi ''$. If $(M_ i, \varphi _ i)_{i \in I}$ is a totally ordered collection of such pairs, then we obtain a map $\bigcup _{i \in I} M_ i \to J$ defined by $a \in M_ i$ maps to $\varphi _ i(a)$. Thus Zorn's lemma applies. To conclude we have to show that if the pair $(M', \varphi ')$ is maximal then $M' = N$. In other words, it suffices to show, given any subgroup $M \subset N$, $M \not= N$ and any $\varphi : M \to J$, then we can find $\varphi ' : M' \to J$ with $M \subset M' \subset N$ such that (a) the inclusion $M \subset M'$ is strict, and (b) the morphism $\varphi '$ extends $\varphi $.

To prove this, pick $x \in N$, $x \not\in M$. Let $I = \{ f \in R \mid fx \in M\} $. This is an ideal of $R$. Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi (fx)$. Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3). By our choice of $I$ the kernel of $M \oplus R \to J$, $(y, f) \mapsto \varphi (y) + \tilde\psi (f)$ contains the kernel of the map $M \oplus R \to N$, $(y, f) \mapsto y + fx$. Hence this homomorphism factors through the image $M' = M + Rx$ and this extends the given homomorphism as desired. $\square$